Roboguru

Diketahui segitiga  dengan titik sudut masing-masing ,  dan . Tentukan: b. Gambar segitiga  dan bayangan segitiga  pada rotasi  berlawanan arah jarum jam dengan pusat rotasi .

Pertanyaan

Diketahui segitiga ABC dengan titik sudut masing-masing straight A open parentheses 2 comma space 1 close parenthesesstraight B open parentheses 4 comma space 2 close parentheses dan straight C open parentheses 2 comma space 4 close parentheses. Tentukan:

b. Gambar segitiga ABC dan bayangan segitiga ABC pada rotasi 30 degree berlawanan arah jarum jam dengan pusat rotasi straight O open parentheses 0 comma space 0 close parentheses.

Pembahasan Soal:

Rotasi terhadap titik asal straight O open parentheses 0 comma space 0 close parentheses, dengan sudut rotasi theta, dapat dihitung dengan rumus berikut.

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row cell cos space theta end cell cell negative sin space theta end cell row cell sin space theta end cell cell cos space theta end cell end table close parentheses open parentheses table row x row y end table close parentheses

Diketahui segitiga ABC dengan titik sudut masing-masing straight A open parentheses 2 comma space 1 close parenthesesstraight B open parentheses 4 comma space 2 close parentheses dan straight C open parentheses 2 comma space 4 close parentheses.

Akan ditentukan Gambar segitiga ABC dan bayangan segitiga ABC pada rotasi 30 degree berlawanan arah jarum jam dengan pusat rotasi straight O open parentheses 0 comma space 0 close parentheses. Karena rotasi berlawanan arah jarum jam, maka Rotasi bernilai positif.

Menentukan bayangan titik straight A open parentheses 2 comma space 1 close parentheses.

straight A open parentheses 2 comma space 1 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight A apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space open parentheses 30 degree close parentheses end cell cell negative sin space open parentheses 30 degree close parentheses end cell row cell sin space open parentheses 30 degree close parentheses end cell cell cos space open parentheses 30 degree close parentheses end cell end table close parentheses open parentheses table row 2 row 1 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row 2 row 1 end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell square root of 3 minus 1 half end cell row cell 1 plus 1 half square root of 3 end cell end table close parentheses end cell end table straight A open parentheses 2 comma space 1 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight A apostrophe open parentheses square root of 3 minus 1 half comma space 1 plus 1 half square root of 3 close parentheses

Menentukan bayangan titik straight B open parentheses 4 comma space 2 close parentheses.

straight B open parentheses 4 comma space 2 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight B apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space open parentheses 30 degree close parentheses end cell cell negative sin space open parentheses 30 degree close parentheses end cell row cell sin space open parentheses 30 degree close parentheses end cell cell cos space open parentheses 30 degree close parentheses end cell end table close parentheses open parentheses table row 4 row 2 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row 4 row 2 end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 2 square root of 3 minus 1 end cell row cell 2 plus square root of 3 end cell end table close parentheses end cell end table straight B open parentheses 4 comma space 2 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight B apostrophe open parentheses 2 square root of 3 minus 1 comma space 2 plus square root of 3 close parentheses

Menentukan bayangan titik straight C open parentheses 2 comma space 4 close parentheses.

straight C open parentheses 2 comma space 4 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight C apostrophe open parentheses x apostrophe comma space y apostrophe close parentheses table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos space open parentheses 30 degree close parentheses end cell cell negative sin space open parentheses 30 degree close parentheses end cell row cell sin space open parentheses 30 degree close parentheses end cell cell cos space open parentheses 30 degree close parentheses end cell end table close parentheses open parentheses table row 2 row 4 end table close parentheses end cell row blank equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row 2 row 4 end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell square root of 3 minus 2 end cell row cell 1 plus 2 square root of 3 end cell end table close parentheses end cell end table straight C open parentheses 2 comma space 4 close parentheses rightwards arrow with open square brackets straight O comma space straight R open parentheses 30 degree close parentheses close square brackets on top straight C apostrophe open parentheses square root of 3 minus 2 comma space 1 plus 2 square root of 3 close parentheses

Jadi, diperoleh titik koordinat bayangan segitiga ABC adalah straight A apostrophe open parentheses 1 comma space minus 2 close parenthesesstraight B apostrophe open parentheses 2 comma space minus 4 close parentheses dan straight C apostrophe open parentheses 4 comma space minus 2 close parentheses. Sehingga gambarnya dapat dibuat seperti berikut.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui lingkaran L berpusat di titik  dan melalui titik . Jika lingkaran L diputar  terhadap titik searah jarum jam, kemudian digeser ke bawah sejauh satuan, maka persamaan lingkaran L yang dihasil...

Pembahasan Soal:

Persamaan transformasi dengan pusat straight O left parenthesis 0 comma space 0 right parenthesis dan sudut rotasi theta sarah jarum jam, ditulis straight R left square bracket 0 comma space minus theta right square bracket, untuk pemetaan left parenthesis x comma space y right parenthesis ke left parenthesis x apostrophe comma space y apostrophe right parenthesis dapat dinyatakan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos theta end cell cell sin theta end cell row cell negative sin theta end cell cell cos theta end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell end table  

  • Sehingga matriks rotasi straight R left square bracket 0 comma space minus 90 degree right square bracket adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos 90 degree end cell cell sin 90 degree end cell row cell negative sin 90 degree end cell cell cos 90 degree end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses open parentheses table row x row y end table close parentheses end cell end table        

Jika T equals open parentheses table row a row b end table close parentheses dan titik benda A open parentheses table row x row y end table close parentheses, maka dapat ditulis sebagai berikut.

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row x row y end table close parentheses plus open parentheses table row a row b end table close parentheses 

  • Sehingga matriks translasi (digeser ke bawah sejauh 5 spacesatuan).

open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses equals open parentheses table row x row y end table close parentheses plus open parentheses table row 0 row cell negative 5 end cell end table close parentheses 

Diketahui lingkaran L berpusat di titik left parenthesis negative 2 comma space 3 right parenthesis dan melalui titik left parenthesis 1 comma space 5 right parenthesis. Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell r squared end cell equals cell open parentheses 1 minus open parentheses negative 2 close parentheses close parentheses squared plus open parentheses 5 minus 3 close parentheses squared end cell row blank equals cell 3 squared plus 2 squared end cell row blank equals cell 9 plus 4 end cell row blank equals 13 end table  

Berdasarkan aturan diatas, maka hasil rotasinya titik pusat left parenthesis negative 2 comma space 3 right parenthesis:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 1 row cell negative 1 end cell 0 end table close parentheses open parentheses table row cell negative 2 end cell row 3 end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 0 times open parentheses negative 2 close parentheses plus 1 times 3 end cell row cell negative 1 times open parentheses negative 2 close parentheses plus 0 times 3 end cell end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 3 row 2 end table close parentheses end cell end table 

Kemudian, ditranslasi:

open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses equals open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses plus open parentheses table row a row b end table close parentheses open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses equals open parentheses table row 3 row 2 end table close parentheses plus open parentheses table row 0 row cell negative 5 end cell end table close parentheses open parentheses table row cell x apostrophe apostrophe end cell row cell y apostrophe apostrophe end cell end table close parentheses equals open parentheses table row 3 row cell negative 3 end cell end table close parentheses 

Sehingga, diperoleh hasil transformasi titik pusat lingkaran L adalah open parentheses 3 comma space minus 3 close parentheses dan melalui titik left parenthesis 1 comma space 5 right parenthesis.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x minus x apostrophe apostrophe close parentheses squared plus open parentheses y minus y apostrophe apostrophe close parentheses squared end cell equals cell r squared end cell row cell open parentheses x minus 3 close parentheses squared plus open parentheses y minus open parentheses negative 3 close parentheses close parentheses squared end cell equals 13 row cell open parentheses x minus 3 close parentheses squared plus open parentheses y plus 3 close parentheses squared end cell equals 13 row cell x squared minus 6 x plus 9 plus y squared plus 6 y plus 9 end cell equals 13 row cell x squared minus 6 x plus 9 plus y squared plus 6 y plus 9 minus 13 end cell equals 0 row cell x squared plus y squared minus 6 x plus 6 y plus 5 end cell equals 0 end table   

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Pada gambar di bawah ini tuliskan hasil bangun perputaran terhadap titik  pada: c. sumbu  dengan putaran sejauh  searah jarum jam.

Pembahasan Soal:

Jadi, sumbu begin mathsize 14px style OE end style dengan putaran sejauh begin mathsize 14px style 240 degree end style searah jarum jam di gambarkan seperti di atas.

0

Roboguru

Tentukan bayangan dari garis   dirotasikan 90° dengan pusat O(0,0)!

Pembahasan Soal:

Asumsikan rotasi berlawanan dengan arah jarum jam. Maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell cos open parentheses 90 degree close parentheses end cell cell negative sin open parentheses 90 degree close parentheses end cell row cell sin open parentheses 90 degree close parentheses end cell cell cos open parentheses 90 degree close parentheses end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row blank equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row x row y end table close parentheses end cell row blank equals cell open parentheses table row cell negative y end cell row x end table close parentheses end cell end table

Maka, y equals negative x apostrophe dan x equals y apostrophe.

Sehingga diperoleh persamaan bayangan garisnya adalah

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 3 x plus 2 end cell row cell negative x apostrophe end cell equals cell 3 y apostrophe plus 2 end cell row cell 3 y apostrophe plus x apostrophe plus 2 end cell equals 0 row cell x plus 3 y plus 2 end cell equals 0 end table 

0

Roboguru

Bayangan parabola  jika diputar  berlawanan arah dengan jarum jam dan pusat  ....

Pembahasan Soal:

Persamaan transformasi dengan pusat straight O left parenthesis 0 comma space 0 right parenthesis dan sudut rotasi theta berlawanan arah jarum jam, ditulis straight R left square bracket 0 comma space theta right square bracket , untuk pemetaan left parenthesis x comma space y right parenthesis ke left parenthesis x apostrophe comma space y apostrophe right parenthesis dapat dinyatakan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell R subscript theta end cell equals cell open parentheses table row cell cos theta end cell cell negative sin theta end cell row cell sin theta end cell cell cos theta end cell end table close parentheses end cell end table 

Sehingga matriks rotasi straight R left square bracket 0 comma space 30 degree right square bracket adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell R subscript 30 degree end subscript end cell equals cell open parentheses table row cell cos 30 degree end cell cell negative sin 30 degree end cell row cell sin 30 degree end cell cell cos 30 degree end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses end cell end table     

Maka, rotasi oleh straight R left square bracket 0 comma space 30 degree right square bracket adalah sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 3 end cell cell negative 1 half end cell row cell 1 half end cell cell 1 half square root of 3 end cell end table close parentheses to the power of negative 1 end exponent open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 1 half end style square root of 3 times begin display style 1 half end style square root of 3 minus open parentheses negative begin display style 1 half end style close parentheses times begin display style 1 half end style end fraction open parentheses table row cell 1 half square root of 3 end cell cell 1 half end cell row cell negative 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 3 over 4 end style plus begin display style 1 fourth end style end fraction open parentheses table row cell 1 half square root of 3 end cell cell 1 half end cell row cell negative 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell fraction numerator 1 over denominator begin display style 1 end style end fraction open parentheses table row cell 1 half square root of 3 end cell cell 1 half end cell row cell negative 1 half end cell cell 1 half square root of 3 end cell end table close parentheses open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell row cell open parentheses table row x row y end table close parentheses end cell equals cell open parentheses table row cell 1 half square root of 3 x apostrophe plus 1 half y apostrophe end cell row cell negative 1 half x apostrophe plus 1 half square root of 3 y apostrophe end cell end table close parentheses end cell end table end style    

Dari kesamaan matriks di atas, diperoleh:

  x equals 1 half square root of 3 x apostrophe plus 1 half y apostrophe y equals negative 1 half x apostrophe plus 1 half square root of 3 y apostrophe 

Substitusikan nilai x dan y yang diperoleh ke persamaan awal.

  table attributes columnalign right center left columnspacing 0px end attributes row y equals cell x squared plus 1 end cell row cell open parentheses negative 1 half x apostrophe plus 1 half square root of 3 y apostrophe close parentheses end cell equals cell open parentheses 1 half square root of 3 x apostrophe plus 1 half y apostrophe close parentheses squared plus 1 end cell end table 

Sehingga diperoleh persamaan bayangannya adalah open parentheses negative 1 half x plus 1 half square root of 3 y close parentheses equals open parentheses 1 half square root of 3 x plus 1 half y close parentheses squared plus 1

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Diketahui garis  dicerminkan terhadap sumbu y dan diputar , maka persamaan bayangan garis tersebut adalah ....

Pembahasan Soal:

Matriks refleksi tehadap sumbu y adalah sebagai berikut.

T subscript 1 equals open parentheses table row cell negative 1 end cell 0 row 0 1 end table close parentheses  

Matriks rotasi straight R left square bracket straight O comma space 90 degree right square bracket adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell T subscript 2 end cell equals cell open parentheses table row cell cos 90 degree end cell cell negative sin 90 degree end cell row cell sin 90 degree end cell cell cos 90 degree end cell end table close parentheses end cell row blank equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses end cell end table         

 Maka, transformasi oleh T subscript 1 dilanjutkan T subscript 2 adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell T subscript 2 ring operator T subscript 1 open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 cell negative 1 end cell row 1 0 end table close parentheses open parentheses table row cell negative 1 end cell 0 row 0 1 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 0 times open parentheses negative 1 close parentheses plus open parentheses negative 1 close parentheses times 0 end cell cell 0 plus open parentheses negative 1 close parentheses times 1 end cell row cell 1 times open parentheses negative 1 close parentheses plus 0 end cell cell 1 times 0 plus 0 times 1 end cell end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row 0 cell negative 1 end cell row cell negative 1 end cell 0 end table close parentheses open parentheses table row x row y end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell 0 times x plus open parentheses negative 1 close parentheses times y end cell row cell open parentheses negative 1 close parentheses times x plus 0 times y end cell end table close parentheses end cell row cell open parentheses table row cell x apostrophe end cell row cell y apostrophe end cell end table close parentheses end cell equals cell open parentheses table row cell negative y end cell row cell negative x end cell end table close parentheses end cell end table    

Dari kesamaan matriks di atas, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell x apostrophe end cell equals cell negative y space left right arrow space y equals negative x apostrophe end cell row cell y apostrophe end cell equals cell negative x space left right arrow space x equals negative y apostrophe end cell end table   

Substitusikan nilai x dan y yang diperoleh ke persamaan awal.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus y minus 4 end cell equals 0 row cell 2 open parentheses negative y apostrophe close parentheses minus open parentheses negative x apostrophe close parentheses minus 4 end cell equals 0 row cell negative 2 y apostrophe plus x apostrophe minus 4 end cell equals 0 row cell x apostrophe minus 2 y apostrophe minus 4 end cell equals 0 end table 

Sehingga, diperoleh bayangannya adalah x minus 2 y minus 4 equals 0 

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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